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I recently encounter an integral problem consisting the integral $$ I:=\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y, $$ I tried to tackle $I$ using the double angle formula and the result of $$ \int_{0}^{\frac{\pi}{4}} y\ln (\cos y) d y $$

\begin{aligned} I &=\int_{0}^{\frac{\pi}{2}} y \ln \left(2 \cos ^{2} \frac{y}{2}\right) d y \\ &=\ln 2 \int_{0}^{\frac{\pi}{2}} y d y+2 \int_{0}^{\frac{\pi}{2}} y \ln \left(\cos \frac{y}{2}\right) d y \\ &=\frac{\pi^{2}}{8} \ln 2+8 \int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y \end{aligned}

By my post,$$\int_{0}^{\frac{\pi}{4}} y\ln (\cos y) d y = \frac{\pi G}{8}-\frac{\pi^{2}}{32} \ln 2-\frac{21}{128} \zeta(3) $$

Now we can conclude that $$ \boxed{\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y = \pi G-\frac{21}{16} \zeta(3)-\frac{\pi^{2}}{8} \ln 2} $$

Suggestions for improvement and alternative methods are warmly welcome!

Lai
  • 20,421

2 Answers2

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Another solution, obtained by one integration by parts is $$\int y \log (1+\cos (y))\,dy=$$ $$\frac{1}{2} y^2 \log (1+\cos (y))+2 i y \text{Li}_2\left(-e^{i y}\right)-2 \text{Li}_3\left(-e^{i y}\right)+\frac{i y^3}{6}-y^2 \log \left(1+e^{i y}\right)$$ giving for the definte integral $$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy=\pi C-2 \text{Li}_3(-i)-\frac{1}{4} \pi ^2 \log (1+i)-\frac{3 }{2} \zeta (3)$$ $$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy=\pi C-\frac{21 }{16}\zeta (3)-\frac{1}{8} \pi ^2 \log (2)$$

Just for the fun

Using the $1,400$ years old approximation $$\cos(y) \simeq\frac{\pi ^2-4y^2}{\pi ^2+y^2}\qquad \text{for} \qquad -\frac \pi 2 \leq y\leq\frac \pi 2$$ we should have $$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy\sim \pi ^2 \log \left(\frac{4}{5^{5/6}}\right) $$ (relative error equal to $0.07$%).

Claude Leibovici
  • 260,315
1

Using the series expansion of $ \log (\cos (y))$ an alternative method could be $$\int y \log (\cos (y))\,dy=\int\sum_{n=1}^\infty (-1)^n \,\frac{2^{2(n-1)}}{(2 n)!}\big[E_{2 n-1}(1)-E_{2 n-1}(0)\big]\, y^{2n+1}\,dy$$ where appear Euler polynomials.

So, $$\int_0^{\frac \pi 4} y \log (\cos (y))\,dy=\sum_{n=1}^\infty (-1)^n \,\frac{ \pi ^{2( n+1)}}{2^{2 n+7}\,(n+1)\, \Gamma (2 n+1)}\big[E_{2 n-1}(1)-E_{2 n-1}(0)\big]$$

Edit

We could do something similar using

$$\log (1+\cos (y))=\log(2)+\sum_{n=1}^\infty (-1)^n \,\Bigg[\frac{E_{2 n}(-1)-E_{2 n}(1)}{2 (2 n)!}+\frac{E_{2 n-1}(-1)+E_{2 n-1}(1)}{4 n (2 n-1)!}\Big]y^{2n}$$

Claude Leibovici
  • 260,315